`x :3*5 = 3/4 :(-5/6)`

`x :15 =3/4*(-6/5)=-9/10`

`x = -9/10 *15 =-27/2`

`x-1*2/2 = 8/x -1.2`

`x- 1*1 = 8/x -2`

`x-8/x = -2+1`

`x-8/x =-1`

`x^2 -8x =-x`

`x^2 -8x +x=0`

`x^2 -7x =0`

`x(x-7) =0`

`=>[(x=0),(x=7):}`

`a, x \div 15=-9/10`

`x=-9/10*14`

`x=-27/2`

`b, (x-1*2)/2=8/(x-1*2)`

\(\left(x-1\cdot2\right)\cdot\left(x-1\cdot2\right)=8\cdot2\)

`(x-1*2)^2=16`

`(x-1*2)^2=(+-4)^2`

\(\Rightarrow\left[{}\begin{matrix}x-1\cdot2=4\\x-1\cdot2=-4\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4+2\\x=\left(-4\right)+2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

b: \(5x^2+3x-2-4x^2+x+5=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

a) \(\Rightarrow x^8-2x^4-8=0\Rightarrow\left(x^4-4\right)\left(x^4+2\right)=0\)

\(\Rightarrow\left(x^2-2\right)\left(x^2+2\right)\left(x^4+2\right)=0\)

\(\Rightarrow x^2=2\Rightarrow x=\pm\sqrt{2}\)(do \(x^2+2\ge2>0,x^4+2\ge2>0\))

b) \(\Rightarrow x^2+4x+3=0\Rightarrow\left(x+1\right)\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

\(a,\Leftrightarrow x^8-2x^4-8=0\\ \Leftrightarrow x^8+2x^4-4x^4-8=0\\ \Leftrightarrow\left(x^4+2\right)\left(x^4-4\right)=0\\ \Leftrightarrow\left(x^4+2\right)\left(x^2-2\right)\left(x^2+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\left(x^2+2>0;x^4+2>0\right)\\ b,\Leftrightarrow x^2+4x+3=0\\ \Leftrightarrow\left(x+1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

\(a,\Rightarrow x=19-17=2\\ b,\Rightarrow x+8=28:2=14\\ \Rightarrow x=14-8=6\\ c,\Rightarrow42-x=5^2=25\\ \Rightarrow x=42-25=17\)

a)x=2

b)x+8=14

x=6

c)\(42-x=5^2\)

\(42-x=25\)

\(-x=-17\)

\(x=17\)

a)x+17=19

⇔x=2

b)2(x+8)=28

⇔x+8=14

⇔x=6

c)42-x=5⁸:5⁶

⇔42-x=5²

⇔42-x=25

⇔x=17

`a)x^3=343=7^3`

`=>x=7`

Vậy `x=7`

`b)(x-2,5)^4=(x-2,5)^2`

`=>(x-2,5)^2[(x-2,5)^2-1]=0`

`+)(x-2,5)^2=0<=>x=2,5`

`+)(x-2,5)^2=1`

`TH1:x-2,5=1<=>x=3,5`

`th2:x-2,5=-1<=>x=1,5`

Vậy `x=0` hoặc `x=1,5` hoặc `x=3,5

`c)x^8/243=27`

`=>x^8=27.243`

`=>x^8=3^3*3^5=3^8`

`=>x=+-3`

Giải:

a) \(x^3=343\) 

\(\Rightarrow x^3=7^3\) 

\(\Rightarrow x=7\) 

b) \(\left(x-2,5\right)^4=\left(x-2,5\right)^2\) 

Vì có cùng cơ số nhưng lại khác mũ thì \(\left(x-2,5\right)\in\left\{0;1\right\}\) 

TH1:

\(\Rightarrow x-2,5=0\) 

\(\Rightarrow x=2,5\) 

TH2:

\(\Rightarrow\left[{}\begin{matrix}x-2,5=1\\x-2,5=-1\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=3,5\\x=1,5\end{matrix}\right.\) 

c) \(\dfrac{x^8}{243}=27\) 

\(\Rightarrow x^8=3^3.3^5\) 

\(\Rightarrow x^8=3^8\) 

\(\Rightarrow x=\pm3\)

a) 2(2x - 8) = 24

2x - 8 = 12

2x = 20

x = 10

b)3.(2^x - 26) = 18

2^x - 26 = 6

2^x = 32

x = 5

a) \(3\left(x-2\right)+2\left(x-3\right)=5\)

\(\Rightarrow3x-6+2x-6=5\)

\(\Rightarrow5x=17\Rightarrow x=\dfrac{17}{5}\)

b) \(\left(2x-8\right)^2-16=0\)

\(\Rightarrow\left(2x-8-4\right)\left(2x-8+4\right)=0\)

\(\Rightarrow\left(2x-12\right)\left(2x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x=12\\2x=4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)

c) \(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=3\)

\(\Rightarrow4x^2-4x+1-4x^2+12x-x+3=3\)

\(\Rightarrow7x=-1\Rightarrow x=-\dfrac{1}{7}\)

a: Ta có: \(3\left(x-2\right)+2\left(x-3\right)=5\)

\(\Leftrightarrow3x-6+2x-6=5\)

\(\Leftrightarrow5x=17\)

hay \(x=\dfrac{17}{5}\)

b: Ta có: \(\left(2x-8\right)^2-16=0\)

\(\Leftrightarrow\left(2x-4\right)\left(2x-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

a. \(3\left(x-2\right)+2\left(x-3\right)=5\)

\(\Leftrightarrow3x-6+2x-6=5\)

\(\Leftrightarrow5x=17\)

\(\Leftrightarrow x=\dfrac{17}{5}\)

b. \(\left(2x-8\right)^2-16=0\)

\(\Leftrightarrow\left(2x-8-4\right)\left(2x-8+4\right)=0\)

\(\Leftrightarrow4\left(x-6\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)

c. \(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=3\)

\(\Leftrightarrow4x^2-4x+1-4x^2+11x+3-3=0\)

\(\Leftrightarrow7x+1=0\)

\(\Leftrightarrow x=-\dfrac{1}{7}\)

a) Ta có: 12-5x=37

\(\Leftrightarrow5x=-25\)

hay x=-5

Vậy: x=-5

b) Ta có: 7-3|x-2|=-11

\(\Leftrightarrow3\left|x-2\right|=18\)

\(\Leftrightarrow\left|x-2\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{8;-4\right\}\)

c) Ta có: \(x+\dfrac{2}{8}=-\dfrac{15}{4}\)

\(\Leftrightarrow x=\dfrac{-15}{4}-\dfrac{2}{8}=\dfrac{-15}{4}-\dfrac{1}{4}\)

hay x=-4

Vậy: x=-4

a, \(\Leftrightarrow5x=12-37=-25\)

\(\Leftrightarrow x=-\dfrac{25}{5}=-5\)

Vậy ...

b, \(\Leftrightarrow3\left|x-2\right|=7+11=18\)

\(\Leftrightarrow\left|x-2\right|=\dfrac{18}{3}=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)

Vậy ...

c, \(\Leftrightarrow x=-\dfrac{15}{4}-\dfrac{2}{8}=-4\)

Vậy ..

a)-5; b)8 hoặc -4;c)-4 nhé. Bn tham khảo nhé

b: \(x\in\left\{1;2;3;6\right\}\)

Lời giải:

a. $8-3x=(-7)^2:(-7)=(-7)$

$\Rightarrow 3x=8-(-7)=15$

$\Rightarrow x=15:3=5$

b.

$18\vdots x, 24\vdots x$ nên $x\in ƯC(18,24)$

$\Rightarrow ƯCLN(18,24)\vdots x$

Hay $6\vdots x$

$\Rightarrow x\in\left\{\pm 1; \pm 2; \pm 3; \pm 6\right\}$